Jacobi fields are the null solutions of the index form with fixed endpoints

Let $J$ be a smooth Jacobi field with $J(a)=J(b)=0$, and let $W\in {\mathcal{X}}_0^2(\gamma )$. The integration-by-parts formula [F2] has no derivative jumps for $J$, its outer endpoint term vanishes because $W$ has zero endpoint values, and its interior residual vanishes by [F3]; hence $I_{\gamma }(J,W)=0$, proving the forward inclusion in the fixed-endpoint radical.

Let $V\in {\mathcal{X}}_0^2(\gamma )$ lie in the radical, and on each smooth piece set $A_V:=D_t^{2}V+R(V,\dot{\gamma })\dot{\gamma }$. If $A_V(t_0)\mathrm{\ne }0$ at an interior point, choose a parallel field $E$ with $E(t_0)=A_V(t_0)$ by [F5]; continuity gives $[u,v]$ strictly inside that piece with $g(A_V,E)>0$. Put $\varphi (t)=(t-u{)}^2(v-t{)}^2$ on $[u,v]$ and zero elsewhere; since $E$ is smooth, $W=\varphi E$ is continuous piecewise-$C^2$ and endpoint-zero. Then $h=\varphi g(A_V,E)$ is continuous, nonnegative, and nonzero. Since $W$ vanishes near the subdivision points and outer endpoints, [F2] gives $I_{\gamma }(V,W)=-{\int }_a^{b}h(t)dt=0$, contradicting [F8]. Thus $A_V=0$ on each open piece.

For a smooth Jacobi field $J$ and arbitrary continuous piecewise-$C^1$ $W$, [F2] reduces to $I_{\gamma }(J,W)=g(D_{t}J(b),W(b))-g(D_{t}J(a),W(a))$. If both derivatives vanish, this is zero for every $W$. Conversely, if it is zero for every $W$, extend $D_{t}J(a)$ to a parallel field $E_a$ and take $W_a(t)=\frac{b-t}{b-a}{E}_a(t)$, giving $0=I_{\gamma }(J,W_a)=-\mathrm{\mid }{D}_{t}J(a){\mathrm{\mid }}^2$; extend $D_{t}J(b)$ to a parallel field $E_b$ and take $W_b(t)=\frac{t-a}{b-a}{E}_b(t)$, giving $0=I_{\gamma }(J,W_b)=\mathrm{\mid }{D}_{t}J(b){\mathrm{\mid }}^2$. Positive definiteness [F9] yields both endpoint derivatives zero.

At an interior breakpoint $t_j$, write ${\mathrm{\Delta }}_j{D}_{t}V=D_{t}V(t_j^{+})-D_{t}V(t_j^{-})$. If it were nonzero, choose a parallel field $E$ with $E(t_j)={\mathrm{\Delta }}_j{D}_{t}V$ by [F5] and a piecewise-linear hat $\eta$ equal to $1$ at $t_j$, zero at its neighboring subdivision points and zero elsewhere. Then $W=\eta E$ is in ${\mathcal{X}}_0^2(\gamma )$ and vanishes at every other breakpoint. The residual is already zero by step 1.2, so [F2] gives $I_{\gamma }(V,W)=-g({\mathrm{\Delta }}_j{D}_{t}V,{\mathrm{\Delta }}_j{D}_{t}V)<0$, contradicting the radical condition. Hence every derivative jump vanishes.

On each smooth piece, local parallel frames exist by extending a finite basis at any time using [F5]; uniqueness in both time directions makes each extension a frame. In that frame the components $x$ satisfy $x^{\mathrm{\prime }\mathrm{\prime }}+B(t)x=0$, with smooth $B$ by [F10] and the parallel rule [F6, F7], so the piecewise-$C^2$ solution is smooth by repeated differentiation. The global initial-data theorem [F4] gives a unique smooth Jacobi field $J_0$ with $J_0(a)=V(a)$ and $D_t{J}_0(a)=D_{t}V(a)$. On the first piece $V$ and $J_0$ solve the same equation with the same data, so uniqueness [F4] makes them equal there; continuity and the zero derivative jump at each breakpoint give matching data on the next piece. Induction over the finite subdivision yields $V=J_0$ throughout, and $V(a)=V(b)=0$. This proves the reverse inclusion.

A supplied segment excludes an empty-manifold instance. In dimension zero every field is zero; in dimension one the same integration-by-parts argument divides by no dimension and assumes no curvature sign. For a constant geodesic the residual equation is $D_t^{2}V=0$, characterized by the same tests. The condition $a<b$ excludes a degenerate interval, included endpoints use one-sided traces [F2], and the sole choice assumption is the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ [A1]; every parallel section is uniquely determined by its individually specified initial vector, so no further choice is made. The fixed-endpoint iff is proved in steps 1.1–3.1, and the unrestricted iff in step 1.3. [A1, F2, F5, F9, step 1.1, step 1.2, step 1.3, step 2.1, step 3.1] $\mathrm{\square }$